4.通项公式为an=4n 2的等差数列的前n项和公式为________.
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A.an=2n-1Sn=n^2Sn-1=(n-1)^2an=Sn-Sn-1=n^2-(n-1)^2=n^2-n^2+2n-1=2n-1
∵Sn=n2-2n+3,a1=2,∴an=Sn-Sn-1=n2-2n+3-[(n-1)2-2(n-1)+3]=2n-3(n>1),∵当n=1时,a1=-1≠2,∴an=2,n=12n−3,n>1.,故
n=1时,a1=s1=1,n≥2时,an=sn-sn-1=n2+n-1-[(n-1)2+n-1-1]=2n,综上an=1 ,n=12n &nbs
(1)∵数列{a[n]}的通项a[n]=n^2[(cosnπ/3)^2-(sinnπ/3)^2],前n项和为S[n]∴a[n]=n^2Cos(2nπ/3)∴S[n]=1^2(-1/2)+2^2(-1/
(1)由n2-5n+4<0,得1<n<4,故数列中有两项为负数;(2)an=n2-5n+4=(n−52)2-94,因此当n=2或3时,an有最小值,最小值为-2.
an=sn-s(n-1)=n^2+1-(n-1)^2-1=2n-1
n(n+1)(2n+1)/6二次数列也可以叫做二阶等差数列因为各项差是等差数列如果你学过组合数比较好求没学过也能求一般用待定系数待定一个三次的多项式
由数列{an}的前n项和为Sn=n2+3n+1,当n=1时,a1=S1=5;当n≥2时,an=Sn-Sn-1=n2+3n+1-[(n-1)2+3(n-1)+1]=2n+2.当n=1时上式不成立.∴an
∵an=1n2+3n+2=1(n+1)(n+2)=1n+1-1n+2,∴Sn=12-13+13-14+…+1n-1n+1=12-1n+1,∵其前n项和为718,∴12-1n+1=718,解得n=8.故
由an+1+Sn=n2+2n①,得an+Sn-1=(n-1)2+2(n-1)(n≥2)②,①-②得,an+1=2n+1(n≥2),an=2n-1(n≥3),又a1=0,a2=3,所以an=0,n=12
an=1/(n+1)(n+3)=1/2*[1/(n+1)-1/(n+3)]所以Sn=1/2*[1/2-1/4+1/3-1/5+……+1/n-1/(n+2)+1/(n+1)-1/(n+3)]=1/2*[
利用作差法即可a(n+1)-a(n)=(n+1)²+λ(n+1)-[n²+λn]=2n+1+λ由已知条件,{an}是递增数列∴2n+1+λ>0恒成立∵2n+1+λ的最小值是2*1+
an=1/(n+1)(n+3)=1/2*[1/(n+1)-1/(n+3)]所以Sn=1/2*[1/2-1/4+1/3-1/5+……+1/n-1/(n+2)+1/(n+1)-1/(n+3)]=1/2*[
Sn=1*2+2*2^2+3*2^3+4*2^4+……+n*2^n给此式左右乘以2得:2Sn=1*2^2+2*2^3+3*2^4+4*2^5+……+(n-1)*2^n+n*2^(n+1)第一个式子减第
当n≥2时,an=Sn-Sn-1=n2-2n-1-[(n-1)2-2(n-1)-1]=2n-3,当n=1时,a1=S1=1-2-1=-2,不适合上式,∴数列{an}的通项公式an=−2,(n=1)2n
当n≥2时,由a1•a2•a3…an=n2①,得a1•a2•a3…an-1=(n-1)2②,①②得an=n2(n−1)2,又a1=1,∴an=1(n=1)n2(n−1)2(n≥2),故答案为:1(n=
Sn=n^2+4nS(n-1)=(n-1)^2+4(n-1)=n^2+2n-3An=S(n)-S(n-1)=2n+3
∵a1+2a2+3a3+…+nan=n2,当n≥2时,a1+2a2+…+(n-1)an-1=(n-1)2两式相减可得,nan=n2-(n-1)2=2n-1(n≥2)n=1时,a1=1适合上式∴an=2
由题意可知:当n为奇数时,n+1为偶数,an+an+1=0,∴S100=(a1+a2)+(a3+a4)+…+(a99+a100)=0故答案为:0
ai=1/2Sn=1/2(n2+3n-2)-anSn-1=1/2((n-1)^2+3(n-1)-2)-an-1相减2an=2n+1+an-1设参数方程求解后:an-4(n+1)+6=(1/2)^(n-